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Now the rst term in line (1) is divisible by 6 by the induction hypothesis P (n), and the last term is obviously divisible by 6. So it remains only to show that the middle term, 3(n2 + n), is divisible by 6.

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. So starting with tehne cleofut rhaagnidnsgidaecwaedehamveics to share maths support resources. 12 + 22 + 32 +All.m. .cc+p rke2so+urc(eks a+re1r)el2ea=sed under an Attribution Non-commerical Share Alike licence.

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Prove the following statements using only axioms or propositions that have been proven before (e.g. you can not use Proposition 2.24 in the proof of Proposition 2.20). 1. Proposition 2.18(iii): For all k ∈ N, k3 + 5k is divisible by 6.

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b) Find all integers n such that n(2n + 1)(7n + 1) are divisible by 12.

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...for n = k, then we can show that the statement must also be true for n = k + 1. Let’s start by assuming that 10k + 2 is divisible by 6, say 10k + 2 = 6q

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So when n = 1 f(n) is divisible by 6. Let us assume that when n = p where p>1 the result is true.

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1. that (n^2 - n) is divisible by 2 for every integer n 2. that (n^3 - n) is divisible by 6 3. that (n^5 - n) is divisible by 30.

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Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

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